CBSE Class 10 Maths Revision Notes Chapter 6

Class 10 Mathematics Revision Notes for Triangles of Chapter 6

Class 10 is one of the most crucial classes. The course is comprehensive and requires consistent practice. Step-by-step preparation allows students to score well in class 10 Mathematics. This starts with practising the questions with proper Class 10 Mathematics Chapter 6 Notes. It does become challenging for students to access and refer to all the study material, essential questions, CBSE Syllabus, etc for Class 10 Mathematics. To simplify this, Extramarks has provided Class 10 Chapter 6 Mathematics Notes which can be referred to by students. The revision notes have been prepared according to the NCERT Books. 

You can also practice the CBSE previous question papers after reviewing the Class 10 Mathematics Chapter 6 Notes. This will help students to score better in class 10 examinations. You can also download the notes for future reference from the Extramarks website. 

Class 10 Maths Revision Notes for Triangles of Chapter 6 

Class 10 Triangles Notes  – An Overview of the Chapter 

Introduction

Remember Class IX where students were taught that if two figures are the same size and shape, they are said to be congruent. In this chapter, students will learn about figures that share the same shape but not necessarily the same size, they are called similar figures. Students will study the similarities between triangles in particular and use this information to provide a concise explanation of the Pythagoras Theorem.

Note- All congruent figures are similar but all similar figures need not be congruent.

Similar Figures

Two polygons of the same number of sides are similar when :

(i) their corresponding angles are equal 

(ii) their corresponding sides are in the same ratio (or proportion)

Similarity of Triangles

The theorem of similarity of triangles states that two triangles are similar if the two situations below are correct: 

• The corresponding angles of both triangles are equal. 
• The corresponding sides of the triangles are in the same proportion or ratio. 

Consider that equiangular triangles are formed when the corresponding angles of two triangles are equal. 

Thales, a prominent Greek mathematician, revealed some essential insight about two equiangular triangles, which is as follows:

• Any two comparable sides of two equiangular triangles have the same ratio.

He is said to have employed a result known as the Basic Proportionality Theorem.

For example: 

Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and

B such that AP = PQ = QD = DR = RB.

Now, through B, draw any line intersecting the arm AY at C.

Also, through point D, draw a line parallel to BC to intersect AC at E. Observe from the constructions that AD|DB = 3|2. 

Measure AE and EC. 

What about AE|EC? Observe that AE|EC is also equal to 3|2. 

Thus, you can see that in ∆ ABC, DE || BC and 

AD| DB = AE|EC

It is due to the Basic Proportionality Theorem. 

Theorems in Similarity 

Similarity of triangles holds three theorems. These are as follows: 

Theorem 1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 

This can be proved in the following manner: 

In triangle ABC, a line parallel to BC intersects AB and AC at D and E, respectively.

Theorem 2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. 

This theorem can be proved by taking a line DE such that AD|DB = AE|EC and assuming that DE is not parallel to BC. 

Theorem 3: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion), and hence the two triangles are similar. 

Two triangles are similar if they meet the AAA (Angle–Angle–Angle) criteria. Taking two triangles, ABC and DEF shows that A = D, B = E, and C = F.

Note: By the angle sum property of a triangle, two angles of one triangle will equal two angles of another triangle. It is, therefore, possible to formulate the AAA similarity criterion as follows:

A triangle is similar to another triangle if two angles of one triangle are equal to two angles of another triangle.

Theorem 4: If in two triangles, the sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal, and hence the two triangles are similar.

This criterion is the SSS (Side–Side–Side) similarity criterion for two triangles.

This theorem can be proved by taking two triangles, ABC and DEF such that AB|DE = BC|EF = CA|FD.

SSS Criterion of Similarity

Side-side similarity criteria state that triangles with the same ratio of sides and angles will be considered similar when their corresponding sides are in the same ratio.

SAS Criterion of Similarity 

According to the side-angle-side similarity criteria, triangles are similar when their sides, including their angles, are in the same proportion.

Basic Proportionality Theorem 

The fundamental proportionality theorem in mathematics asserts that “if a line is drawn parallel to one side of a triangle and intersects the other two sides in distinct spots, then the other two sides are split in the same ratio.”

You can prove the fundamental proportionality theorem in the following manner: 

• Consider the triangle ABC. Draw a line PQ parallel to the side BC of ABC and intersect the sides AB and AC in P and Q, respectively. Thus, according to the basic proportionality theorem, AP/PB = AQ/QC.

Areas of Similar Triangles

The theorem of the area of similar triangles states that “The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

For instance: Consider two triangles – ΔABC and ΔPQR. 

Area of triangle ABC/Area of Triangle QPR = (AB/PQ)^2 = (BC/QR)^2  = (CA/RP)^2.

Proof: We are given two triangles ABC and PQR, such that ∆ ABC ~ ∆ PQR. 

To prove that ar(ABC)| ar(PQR) = (AB\PQ)^2 = (BC|QR)^2 = (CA|RP)^2. 

We draw altitudes AM and PN of the triangles to find the areas of the two triangles. 

ar (ABC) = 1/2 BC × AM and ar (PQR) = 1/2 QR × PN 

So, ar (ABC)|ar (PQR) = ½ BC × AM | ½ QR × PN  

Now, in ∆ ABM and ∆ PQN. 

∠ B = ∠ Q (As ∆ ABC ~ ∆ PQR)

∠ M = ∠ N (Each is 90°)

So, ∆ ABM ~ ∆ PQN (AA similarity criterion)

Therefore, AM/PN = AB/PQ 

Also, ∆ ABC ~ ∆ PQR (Given)

So, AB/PQ = BC/QR = CA/RP 

Therefore, ar(ABC)/ar(PQR) = AB/PQ * AM/PN  = (AB/PQ)^2

Hence Proved. 

Proof of Pythagoras Theorem Using Similarity

Pythagoras’ theorem states, “In a right-angled triangle, the sum of squares of two sides of a right triangle is equal to the square of the hypotenuse of the triangle.”

Right triangles have a hypotenuse whose square equals the sum of both sides’ squares.

The formula is: Hypotenuse^2 = Base^2 + Perpendicular^2

The proof is as follows: 

Consider the right-angle triangle ABC with CB perpendicular, CA hypotenuse, and AB as a base of the triangle. Consider the right triangle, right-angled at B.

Draw BD ⊥ AC

Now,  △ADB ~ △ABC

So, AD/AB = AB/AC

or AD. AC = AB2 ……………(i)

Also, △BDC ~ △ ABC

So, CD/BC = BC/AC

or CD. AC = BC2 ……………(ii)

Adding (i) and (ii),

1. AC + CD. AC = AB2 + BC2

AC(AD + DC) = AB2 + BC2

AC(AC) = AB2 + BC2

⇒ AC2 = AB2 + BC2

Hence, proved. 

Now Let us Use this Theorem to Prove Pythagoras’ Theorem: – repeat 

The converse of Pythagoras Theorem 

Statement: If the length of a triangle is a, b and c and c^2 = a^2 + b^2, then the triangle is a right-angle triangle.

Proof: Construct another triangle, △EGF, such as AC = EG = b and BC = FG = a.

Note: The formula for the right-angle triangle is as follows: a^2+b^2 = c^2. 

In △EGF, by Pythagoras Theorem:

EF^2 = EG^2 + FG^2 = b^2 + a^2 …………(1)

In △ABC, by Pythagoras Theorem:

AB2 = AC2 + BC2 = b^2 + a^2 …………(2)

From equations (1) and (2), we have;

EF^2 = AB^2

EF = AB

⇒ △ ACB ≅ △EGF (By SSS postulate)

⇒ ∠G is a right angle

Thus, △EGF is a right triangle.

Hence, we can say that the converse of the Pythagorean theorem also holds.

Hence Proved.